Solving Polynomials: Hints, Techniques & Solutions

[ts547]

Homework Statement

Solve for x,

225*sin(x)/x^6-225*cos(x)/x^5-90*sin(x)/x^4+15*cos(x)/x^3-5/(2*x^3)=0

Homework Equations

Finding this very complicated to solve, are there any useful hints or techniques we should know about?

The Attempt at a Solution

Have used numerical method using mathematics software and plotted a graph to identify where the function crosses the x-axis. Would prefer a more analytic approach.

Thank you in advance.

 

[Mark44]

Homework Statement

Solve for x,

225*sin(x)/x^6-225*cos(x)/x^5-90*sin(x)/x^4+15*cos(x)/x^3-5/(2*x^3)=0

Homework Equations

Finding this very complicated to solve, are there any useful hints or techniques we should know about?

The Attempt at a Solution

Have used numerical method using mathematics software and plotted a graph to identify where the function crosses the x-axis. Would prefer a more analytic approach.

Thank you in advance.

I think that you are out of luck regarding an analytic solution.

BTW, is this your equation?
$$225\frac{sin(x)}{x^6} - 225 \frac{cos(x)}{x^5} - 90\frac{sin(x)}{x^4} + 15\frac{cos(x)}{x^3} - \frac{5}{2x^3} = 0$$

 

[ts547]

Yeh that's it. I haven't learned how to do the fancy writing yet. I didnt think there would be an easy way of doing this.

 

[gb7nash]

Unless there's some funny trick to recognize here, there's no way to solve this algebraically. It's best to use a numerical technique. i.e. Bisection method, Newton's method, etc.

 

[Mark44]

Yeh that's it. I haven't learned how to do the fancy writing yet. I didnt think there would be an easy way of doing this.

You can see the LaTeX I wrote by clicking the equation.

 

[Apphysicist]

Heck, I managed to simplify it this equation (check work?):

$$(\frac{15}{x^3} - \frac{6}{x})sin(x) + (1 - \frac{15}{x^2})cos(x) = \frac{1}{6}$$

Edit: LaTeX isn't the easiest, heh. Also, I'm not sure that simplification is even very useful.

 

[HallsofIvy]

I hope that you are aware that this is not a matter of "solving polynomials"! The equation you give is not a polynomial.

 

[ts547]

Apphysicist - Haha good simplification. Not very useful I don't think. :) Never mind ill stick with the numerical approach.

HallsofIvy - Ok no its not a polynomial. Didnt know what else to call it at the time. If your so clever help me with this

https://www.physicsforums.com/showthread.php?p=3075732#post3075732

then you can point out technicalities all you want.